`10 mL` of `10^(-6) M HCl` solution is mixed with `90mL H_(2)O`. `pH` will change approximately:

Options

A. By one unit
B. By `0.3` unit
C. By `0.7` unit
D. By `0.1` unit

Explanation

To solve the problem of how the pH changes when mixing 10 mL of \(10^{-6} M\) HCl with 90 mL of water, we can follow these steps:

### Step 1: Calculate the number of millimoles of HCl
We start by calculating the number of millimoles of HCl in the original solution.

\[ \text{Number of millimoles of HCl} = \text{Concentration (M)} \times \text{Volume (L)} \]

Given:
- Concentration of HCl = \(10^{-6} M\)
- Volume of HCl = \(10 mL = 0.01 L\)

\[ \text{Number of millimoles of HCl} = 10^{-6} \times 0.01 = 10^{-8} \text{ moles} = 10^{-5} \text{ millimoles} \]

### Step 2: Determine the total volume after mixing
Next, we find the total volume of the solution after mixing HCl with water.

\[ \text{Total Volume} = \text{Volume of HCl} + \text{Volume of Water} = 10 mL + 90 mL = 100 mL = 0.1 L \]

### Step 3: Calculate the new concentration of H⁺ ions
Now, we calculate the concentration of H⁺ ions after dilution.

\[ \text{Concentration of H}^+ = \frac{\text{Number of millimoles of HCl}}{\text{Total Volume (L)}} \]

\[ \text{Concentration of H}^+ = \frac{10^{-5} \text{ millimoles}}{0.1 \text{ L}} = 10^{-6} \text{ M} \]

### Step 4: Account for the contribution of H⁺ ions from water
At 25°C, pure water contributes \(10^{-7} M\) of H⁺ ions. Therefore, we need to add this to the H⁺ concentration from HCl.

\[ \text{Total H}^+ = 10^{-6} + 10^{-7} = 1.1 \times 10^{-6} \text{ M} \]

### Step 5: Calculate the new pH
Now we can calculate the new pH using the formula:

\[ \text{pH} = -\log[\text{H}^+] \]

Substituting the total concentration of H⁺ ions:

\[ \text{pH} = -\log(1.1 \times 10^{-6}) \]

Using logarithmic properties:

\[ \text{pH} = -\log(1.1) - \log(10^{-6}) \approx -0.04 + 6 = 6.96 \]

### Step 6: Calculate the change in pH
Initially, the pH of the \(10^{-6} M\) HCl solution is:

\[ \text{pH} = 6 \]

Now, the new pH is approximately \(6.96\).

\[ \text{Change in pH} = \text{New pH} - \text{Initial pH} = 6.96 - 6 = 0.96 \]

### Conclusion
The pH will change approximately by \(0.96\) units.

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